4. Cyclic groups
Lemma 4.1. Let G be a group and let Hi , i ∈ I be a collection of
subgroups of G.
Then the intersection
H=
Hi ,
i∈I
is a subgroup of G
Proof. First note that H is non-empty, as the identity belongs to every
Hi . We have to check that H is closed under products and inverses.
Suppose that g and h are in H. Then g and h are in Hi , for all i. But
then hg ∈ Hi for all i, as Hi is closed under products. Thus gh ∈ H.
Similarly as Hi is closed under taking inverses, g −1 ∈ Hi for all i ∈ I.
But then g −1 ∈ H.
Thus H is indeed a subgroup.
D
Definition-Lemma 4.2. Let G be a group and let S be a subset of G.
The subgroup H = (S) generated by S is equal to the smallest
subgroup of G that contains S.
Proof. The only thing to check is that the word smallest makes sense.
Suppose that Hi , i ∈ I is the collection of subgroups that contain S.
By (4.1), the intersection H of the Hi is a subgroup of G.
On the other hand H obviously contains S and it is contained in
each Hi .
Thus H is the smallest subgroup that contains S.
D
Lemma 4.3. Let S be a non-empty subset of G.
Then the subgroup H generated by S is equal to the smallest subset
of G, containing S, that is closed under taking products and inverses.
Proof. Let K be the smallest subset of G, closed under taking products
and inverses.
As H is closed under taking products and inverses, it is clear that
H must contain K. On the other hand, as K is a subgroup of G, K
must contain H.
But then H = K.
D
Definition 4.4. Let G be a group. We say that a subset S of G gen­
erates G, if the smallest subgroup of G that contains S is G itself.
Definition 4.5. Let G be a group. We say that G is cyclic if it is
generated by one element.
Let G = (a) be a cyclic group. By (4.3)
G = { ai | i ∈ Z }.
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MIT OCW: 18.703 Modern Algebra
Prof. James McKernan
Definition 4.6. Let G be a group and let g ∈ G be an element of G.
The order of g is equal to the cardinality of the subgroup generated
by g.
Lemma 4.7. Let G be a finite group and let g ∈ G.
Then the order of g divides the order of G.
Proof. Immediate from Lagrange’s Theorem.
D
Lemma 4.8. Let G be a group of prime order.
Then G is cyclic.
Proof. If the order of G is one, there is nothing to prove. Otherwise
pick an element g of G not equal to the identity. As g is not equal to
the identity, its order is not one. As the order of g divides the order of
G and this is prime, it follows that the order of g is equal to the order
of G.
But then G = (g) and G is cyclic.
D
It is interesting to go back to the problem of classifying groups of
finite order and see how these results change our picture of what is
going on.
Now we know that every group of order 1, 2, 3 and 5 must be cyclic.
Suppose that G has order 4. There are two cases. If G has an element
a of order 4, then G is cyclic.
We get the following group table.
∗
e a a2 a3
e e a a2 a3
a a a 2 a3 e
a2 a2 a3 e a
a3 a3 e a a2
Replacing a2 by b, a3 by c we get
∗
e
a
b
c
e
e
a
b
c
a
a
b
c
e
b
b
c
e
a
c
c
e
a
b
Now suppose that G does not contain any elements of order 4. Since
the order of every element divides 4, the order of every element must
be 1, 2 or 4. On the other hand, the only element of order 1 is the
identity element. Thus if G does not have an element of order 4, then
every element, other than the identity, must have order 2.
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MIT OCW: 18.703 Modern Algebra
Prof. James McKernan
In other words, every element is its own inverse.
∗
e
a
b
c
e a b c
e a b c
a e ?
b
e
c
e
Now ? must in fact be c, simply by a process of elimination. In fact
we must put c somewhere in the row that contains a and we cannot
put it in the last column, as this already contains c. Continuing in this
way, it turns out there is only one way to fill in the whole table
∗
e
a
b
c
e
e
a
b
c
a
a
e
c
b
b
b
c
e
a
c
c
b
a
e
So now we have a complete classification of all finite groups up to
order five (it easy to see that there is a cyclic group of any order; just
take the rotations of a regular n-gon). If the order is not four, then the
only possibility is a cyclic group of that order. Otherwise the order is
four and there are two possibilities.
Either G is cyclic. In this case there are two elements of order 4
(a and a3 ) and one element of order two (a2 ). Otherwise G has three
elements of order two. Note however that G is abelian.
So the first non-abelian group has order six (equal to D3 ).
One reason that cyclic groups are so important, is that any group
G contains lots of cyclic groups, the subgroups generated by the ele­
ments of G. On the other hand, cyclic groups are reasonably easy to
understand. First an easy lemma about the order of an element.
Lemma 4.9. Let G be a group and let g ∈ G be an element of G.
Then the order of g is the smallest positive number k, such that
k
a = e.
Proof. Replacing G by the subgroup (g) generated by g, we might as
well assume that G is cyclic, generated by g.
Suppose that g l = e. I claim that in this case
G = { e, g, g 2 , g 3 , g 4 , . . . , g l−1 }.
Indeed it suffices to show that the set is closed under multiplication
and taking inverses.
Suppose that g i and g j are in the set. Then g i g j = g i+j . If i + j < l
there is nothing to prove. If i + j ≥ l, then use the fact that g l = e to
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MIT OCW: 18.703 Modern Algebra
Prof. James McKernan
rewrite g i+j as g i+j−l . In this case i + j − l > 0 and less than l. So the
set is closed under products.
Given g i , what is its inverse? Well g l−i g i = g l = e. So g l−i is the
inverse of g i . Alternatively we could simply use the fact that H is
finite, to conclude that it must be closed under taking inverses.
Thus |G| ≤ l and in particular |G| ≤ k. In particular if G is infinite,
there is no integer k such that g k = e and the order of g is infinite and
the smallest k such that g k = e is infinity. Thus we may assume that
the order of g is finite.
Suppose that |G| < k. Then there must be some repetitions in the
set
{ e, g, g 2 , g 3, g 4 , . . . , g k−1 }.
Thus g a = g b for some a = b between 0 and k − 1. Suppose that a < b.
Then g b−a = e. But this contradicts the fact that k is the smallest
D
integer such that g k = e.
Lemma 4.10. Let G be a finite group of order n and let g be an element
of G.
Then g n = e.
Proof. We know that g k = e where k is the order of g. But k divides
n. So n = km. But then
g n = g km = (g k )m = em = e.
D
Lemma 4.11. Let G be a cyclic group, generated by a.
Then
(1) G is abelian.
(2) If G is infinite, the elements of G are precisely
. . . a−3 , a−2 , a−1 , e, a, a2 , a3 , . . .
(3) If G is finite, of order n, then the elements of G are precisely
e, a, a2, . . . , an−2 , an−1 ,
and an = e.
Proof. We first prove (1). Suppose that g and h are two elements of G.
As G is generated by a, there are integers m and n such that g = am
and h = an . Then
gh = am an
= am+n
= an+m
= hg.
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MIT OCW: 18.703 Modern Algebra
Prof. James McKernan
Thus G is abelian. Hence (1).
(2) and (3) follow from (4.9).
D
Note that we can easily write down a cyclic group of order n. The
group of rotations of an n-gon forms a cyclic group of order n. Indeed
any rotation may be expressed as a power of a rotation R through
2π/n. On the other hand, Rn = 1.
However there is another way to write down a cyclic group of order
n. Suppose that one takes the integers Z. Look at the subgroup nZ.
Then we get equivalence classes modulo n, the left cosets.
[0], [1], [2], [3], . . . , [n − 1].
I claim that this is a group, with a natural method of addition. In
fact I define
[a] + [b] = [a + b].
in the obvious way. However we need to check that this is well-defined.
The problem is that the notation
[a]
is somewhat ambiguous, in the sense that there are infinitely many
numbers a' such that
[a' ] = [a].
In other words, if the difference a' − a is a multiple of n then a and a'
represent the same equivalence class. For example, suppose that n = 3.
Then [1] = [4] and [5] = [−1]. So there are two ways to calculate
[1] + [5].
One way is to add 1 and 5 and take the equivalence class. [1] + [5] =
[6]. On the other hand we could compute [1] + [5] = [4] + [−1] = [3].
Of course [6] = [3] = [0] so we are okay.
So now suppose that a' is equal to a modulo n and b' is equal to b
modulo n. This means
a' = a + pn
and
b' = b + qn,
where p and q are integers.
Then
a' + b' = (a + pn) + (b + qn) = (a + b) + (p + q)n.
So we are okay
[a + b] = [a' + b' ],
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MIT OCW: 18.703 Modern Algebra
Prof. James McKernan
and addition is well-defined. The set of left cosets with this law of
addition is denote Z/nZ, the integers modulo n. Is this a group? Well
associativity comes for free. As ordinary addition is associative, so is
addition in the integers modulo n.
[0] obviously plays the role of the identity. That is
[a] + [0] = [a + 0] = [a].
Finally inverses obviously exist. Given [a], consider [−a]. Then
[a] + [−a] = [a − a] = [0].
Note that this group is abelian. In fact it is clear that it is generated
by [1], as 1 generates the integers Z.
How about the integers modulo n under multiplication? There is an
obvious choice of multiplication.
[a] · [b] = [a · b].
Once again we need to check that this is well-defined. Exercise left
for the reader.
Do we get a group? Again associativity is easy, and [1] plays the
role of the identity. Unfortunately, inverses don’t exist. For example
[0] does not have an inverse. The obvious thing to do is throw away
zero. But even then there is a problem. For example, take the integers
modulo 4. Then
[2] · [2] = [4] = [0].
So if you throw away [0] then you have to throw away [2]. In fact
given n, you should throw away all those integers that are not coprime
to n, at the very least. In fact this is enough.
Definition-Lemma 4.12. Let n be a positive integer.
The group of units, Un , for the integers modulo n is the subset of
Z/nZ of integers coprime to n, under multiplication.
Proof. We check that Un is a group.
First we need to check that Un is closed under multiplication. Sup­
pose that [a] ∈ Un and [b] ∈ Un . Then a and b are coprime to n.
This means that if a prime p divides n, then it does not divide a or
b. But then p does not divide ab. As this is true for all primes that
divide n, it follows that ab is coprime to n. But then [ab] ∈ Un . Hence
multiplication is well-defined.
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MIT OCW: 18.703 Modern Algebra
Prof. James McKernan
This rule of multiplication is clearly associative. Indeed suppose that
[a], [b] and [c] ∈ Un . Then
([a] · [b]) · [c] = [ab] · c
= [(ab)c]
= [a(bc)]
= [a] · [bc]
= [a] · ([b] · [c]).
So multiplication is associative.
Now 1 is coprime to n. But then [1] ∈ Un and this clearly plays the
role of the identity.
Now suppose that [a] ∈ Un . We need to find an inverse of [a]. We
want an integer b such that
[ab] = 1.
This means that
ab + mn = 1,
for some integer m. But a and n are coprime. So by Euclid’s algorithm,
such integers exist.
D
Definition 4.13. The Euler φ function is the function ϕ(n) which
assigns the order of Un to n.
Lemma 4.14. Let a be any integer, which is coprime to the positive
integer n.
Then aφ(n) = 1 mod n.
Proof. Let g = [a] ∈ Un . By (4.10) g φ(n) = e. But then
[aφ (n)] = [1].
Thus
aφ(n) = 1
mod n.
D
Given this, it would be really nice to have a quick way to compute
ϕ(n).
Lemma 4.15. The Euler ϕ function is multiplicative.
That is, if m and n are coprime positive integers,
ϕ(mn) = ϕ(m)ϕ(n).
Proof. We will prove this later in the course.
D
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MIT OCW: 18.703 Modern Algebra
Prof. James McKernan
Given (4.15), and the fact that any number can be factored, it suffices
to compute ϕ(pk ), where p is prime and k is a positive integer.
Consider first ϕ(p). Well every number between 1 and p − 1 is auto­
matically coprime to p. So ϕ(p) = p − 1.
Theorem 4.16. (Fermat’s Little Theorem) Let a be any integer. Then
ap = a mod p. In particular ap−1 = 1 mod p if a is coprime to p.
Proof. Follows from (4.14).
D
How about ϕ(pk )? Let us do an easy example.
Suppose we take p = 3, k = 2. Then of the eight numbers between
1 and 8, two are multiples of 3, 3 and 6 = 2 · 3. More generally, if a
number between 1 and pk − 1 is not coprime to p, then it is a multiple
of p. But there are pk−1 − 1 such multiples,
p = 1 · p, 2p, 3p, . . . (pk−1 − 1)p.
Thus (pk − 1) − (pk−1 − 1) − pk − pk−1 numbers between 1 and pk are
coprime to p. We have proved
Lemma 4.17. Let p be a prime number. Then
ϕ(pk ) = pk − pk−1 .
Example 4.18. What is the order of U5000 ?
5000 = 5 · 1000 = 5 · (10)3 = 54 · 23 .
Now
ϕ(23 ) = 23 − 22 = 4,
and
ϕ(54 ) = 54 − 53 = 53 (4) = 125 · 4.
As the Euler-phi function is multiplicative, we get
ϕ(5000) = 4 · 4 · 125 = 24 · 53 = 2000.
It is also interesting to see what sort of groups one gets. For example,
what is U6 ?
ϕ(6) = ϕ(2)ϕ(3) = 1 · 2 = 2. Thus we get a cyclic group of order 2.
In fact 1 and 5 are the only numbers coprime to 6.
52 = 25 = 1
mod 6.
How about U8 ? Well
ϕ(8) = 4.
So either U8 is either cyclic of order 4, or every element has order 2.
1, 3, 5 and 7 are the numbers coprime to 2. Now
32 = 9 = 1
mod 8,
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MIT OCW: 18.703 Modern Algebra
Prof. James McKernan
52 = 25 = 1
mod 8,
72 = 49 = 1
mod 8.
and
So
[3]2 = [5]2 = [7]2 = [1]
and every element of U8 , other than the identity, has order two. But
then U8 cannot be cyclic.
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MIT OCW: 18.703 Modern Algebra
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